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Constant Acceleration of Two Cars

  1. Feb 2, 2008 #1
    [SOLVED] Constant Acceleration

    1. The problem statement, all variables and given/known data

    Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h and car B maintains a constant speed of 110 km/hr. At t=0, car B is 45 km behind Car A.
    A) How much farther will car A travel before Car b overtakes it?
    B)How much ahead of A will B be in 30 seconds after it overtakes A?

    2. Relevant equations

    3. The attempt at a solution

    I am stumped. I read and reread the section on my textbook about constant acceleration and the material is so confusing. Can someone please hint me or point me to a direction of understanding this please. Thanks.
  2. jcsd
  3. Feb 2, 2008 #2
    Why constant acceleration? Acceleration is 0 as you only have constant speed right?
  4. Feb 2, 2008 #3

    Doc Al

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    Staff: Mentor

    Come up with expressions for the position of each car as a function of time. I'd call the location of Car b at t = 0 to be the origin. (So Car A is at 45 km at t = 0.)

    There's no acceleration mentioned, so this should be easy.
  5. Feb 2, 2008 #4
    110x = 80x+45
    is this the right approach? Im really lost.
  6. Feb 2, 2008 #5
    Your equation is right, but how did you get x = .5? And what are the units? For that matter, what is x? What have you solved for here?

    If you explain your reasoning a little more, we can help you better.
  7. Feb 2, 2008 #6
    At the time instant where two cars meet up, the distance travelled by Car B will be 45 more than Car A...
    v = x/t is the main equation for this type of question.
  8. Feb 2, 2008 #7

    .5 hr? it will take .5 hr for Car B to catch up to Car A?

    So Car A will travel another 40 km before Car B catches up.
    Last edited: Feb 2, 2008
  9. Feb 2, 2008 #8
    Okay, so x is time (it's more conventional to use "t" for a time interval and "x" for distance, but it's not necessary). I think you just need to be a little more careful in solving your equation for x - you just made an algebra mistake.
  10. Feb 2, 2008 #9
    Oops, thanks, Suppose to be 45/30 = 1.5.
    So Car A will travel another 120 km before Car B catches up?

    B)asks the distance Car B will be ahead of for 30 seconds after it passes Car A.
    Car B 110 km/h(110 km per 3600 seconds), Car A 80 km/hr(80 km per 3600 seconds)
    each second for Car B covers .031km and each second for Car A covers .022km
    .031km * 30sec = .93 km.

    So after 30 seconds, Car B will be .93 km ahead of Car A

    Is this approach correct?
  11. Feb 2, 2008 #10
    Not quite - .031 km/sec * 30 sec gives the distance B has traveled, but A isn't just standing still - it's covered some distance, too. The difference between the two tells you how far B is ahead of A.
  12. Feb 2, 2008 #11
    Cool! Thanks belliott, much obliged.
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